# Can I use l'Hôpital's rule on this?

Question:

$\displaystyle\lim\limits_{x\to\pi/2}\frac{\ln(\sin x)}{\cos x}$

This was in a set of questions about l'Hôpital's rule.

However, the numerator is undefined and the denominator equals $0.5$ at $x$ rather than $0/0$ or $\infty/\infty$. How would I solve this?

• $\cos(\pi/2)=0$. Why 0.5? – Przemysław Scherwentke Nov 20 '14 at 0:30
• Oh what. I didn't have a calculator on hand and google told me it was .5. thanks! P.S. since the top is undefined does that mean I can treat it as 0? Since it approaches 0. – user3362196 Nov 20 '14 at 0:30
• Likewise $\ln \sin (\pi/2) = \ln 1 = 0$. Also, Windows has a calculator, calc.exe. – Graham Kemp Nov 20 '14 at 0:54
• @user3362196 Google told you it was $0.5$? Strange—I typed cos(pi/2) into Google just now, and it gave me $0$. (Perhaps you accidentally typed it cos(pi/3)?) – Akiva Weinberger Nov 20 '14 at 0:55

## 2 Answers

Both the numerator and the denominator are $0$, so it is an indeterminate form. If you are unsure about this, look at the sine and cosine waves.

You can apply L'Hopital's rule to the function:

$$\lim_{x\to \frac{\pi}{2}} \frac{\ln(\sin(x))}{\cos(x)}=\lim_{x\to\frac{\pi}{2}}\frac{\cos x}{\sin x \times -\sin x}=\lim_{x\to \frac{\pi}{2}}\frac{-\cos x}{\sin^2 x}$$.

Can you continue?

Yes,

$$\lim_{x \rightarrow \frac{\pi}{2}}\frac{\ln(\sin(x))}{\cos(x)} = \lim_{x \rightarrow \frac{\pi}{2}}\frac{\cos(x)}{-\sin^2(x)}$$

Therefore

$$\lim_{x \rightarrow \frac{\pi}{2}}-\cot(x) \csc(x) = -\cot(\frac{\pi}{2}) \csc(\frac{\pi}{2})= 0$$

• Hmm are you sure? My prof says that it should be 0. – user3362196 Nov 20 '14 at 0:33
• I did it wrong at first, sorry for the confusion – Eric L Nov 20 '14 at 0:34
• What's the point in using the cosecant and the cotangent? It's just unnecessary complication: $\cos(\pi/2)=0$ and $-\sin^2(\pi/2)=-1$. – egreg Nov 20 '14 at 0:38
• @egreg What do you mean? $$\frac{\cos(x)}{-\sin^2(x)} = -\cot(x) \csc(x)$$ – Eric L Nov 20 '14 at 0:38
• I have a habit of expanding things out like so, its another perspective and it doesn't hurt. – Eric L Nov 20 '14 at 0:41